2 H₂O₂ --> 2 H₂ + 2 O₂
2 moles H₂O₂ ------> 2 moles O₂
8 moles H₂O₂ ------> ?
moles O₂ = 8 x 2 / 2
moles O₂ = 16 / 2
= 8 moles
Answer C
hope this helps!
According to the question, the determined melting point of the compound is 112.5-113.0oC. When the solidified compound was retried, the melting point was found to be 133.6-154.5oC. This greater range higher than 112°C is caused by reusing samples leads to errors.
A pure sample is known by its sharp melting point. A pure sample does not melt over a large range. We can see this in the predetermined melting points of the pure sample(112.5-113.0oC).
However, reusing a sample introduces errors because the pure sample may become contaminated leading to a larger and higher range of melting point (133.6-154.5oC) which is far above 112°C.
Learn more: brainly.com/question/5325004
When sulfate (SO₄²⁻) serves as the electron acceptor at the end of a respiratory electron transport chain, the product is hydrogen sulfide (H₂S).
How sulfate acts as electon acceptor and electron donor?
- Sulfate (SO₄²⁻) is used as the electron acceptor in sulfate reduction, which results in the production of hydrogen sulfide (H2S) as a metabolic byproduct.
- Many Gram negative bacteria identified in the -Proteobacteria use sulfate reduction, which is a rather energy-poor process.
- Gram-positive organisms connected to Desulfotomaculum or the archaeon Archaeoglobus also utilise it.
- Electron donors are needed for sulfate reduction, such as hydrogen gas or the carbon molecules lactate and pyruvate (organotrophic reducers) (lithotrophic reducers).
Learn more about the Electron transport chain with the help of the given link:
brainly.com/question/24372542
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Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,
pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
![\frac{[Deprotonated]}{[Protonated]}=0.01](https://tex.z-dn.net/?f=%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D%3D0.01)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
![\frac{[Protonated]}{[Deprotonated]}=100](https://tex.z-dn.net/?f=%5Cfrac%7B%5BProtonated%5D%7D%7B%5BDeprotonated%5D%7D%3D100)
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
